General form of circle — converting, finding centre and radius: Question bank

Before we start, one reminder of the machinery we lean on everywhere below. For $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ the ==centre is $(- g, - f)$ and the radius is $r = g^{2} + f^{2} - c$ ==. The quantity $g^{2} + f^{2} - c$ is the star of the show: it is $r^{2}$ , so its sign decides whether we have a real circle, a single point, or nothing real at all.

The three pictures below are your anchors for the whole page — refer back to them as you work.

Figure — General form of circle — converting, finding centre and radius

A quick worked completion of the square, so nothing below is hand-waving. Starting from $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ :

$group x (x^{2} + 2 g x) + group y (y^{2} + 2 f y) + c = 0$

$x^{2} + 2 g x = (x + g)^{2} - g^{2}, y^{2} + 2 f y = (y + f)^{2} - f^{2}$

$(x + g)^{2} - g^{2} + (y + f)^{2} - f^{2} + c = 0 ⟹ (x + g)^{2} + (y + f)^{2} = g^{2} + f^{2} - c .$

Comparing with the standard form $(x - h)^{2} + (y - k)^{2} = r^{2}$ from Standard form of circle gives $h = - g$ , $k = - f$ , and $r^{2} = g^{2} + f^{2} - c$ . That is the entire toolkit — every trap below is a failure to respect one of these three matched pieces. The folding move itself is Completing the square.

True or false — justify

x^{2} + y^{2} + 6 x - 4 y - 12 = 0

has its centre at

(6, - 4)

False — centre is

(- g, - f)

, and here

2 g = 6, 2 f = - 4

g = 3, f = - 2

, giving centre

(- 3, 2)

. You must negate the halved coefficients (Figure 1 shows why).

The constant term

c

equals

r^{2}

False — from the completion above,

c = h^{2} + k^{2} - r^{2}

, so it mixes centre position with radius. The radius comes from

r^{2} = g^{2} + f^{2} - c

, never from

c

alone.

Any equation of the form

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

is a genuine circle.

False — it is only a real circle when

g^{2} + f^{2} - c > 0

. If that quantity is

0

it is a single point, and if negative there are no real points at all (Figure 2).

The equation

2 x^{2} + 2 y^{2} + 4 x - 8 y + 2 = 0

has centre

(- 2, 4)

False — the

x^{2}, y^{2}

coefficients must be

1

first. Divide by

2

to get

x^{2} + y^{2} + 2 x - 4 y + 1 = 0

, so

g = 1, f = - 2

and the centre is

(- 1, 2)

g^{2} + f^{2} - c = 0

the equation has no solutions.

False — it has exactly one solution, the point

(- g, - f)

, because

(x + g)^{2} + (y + f)^{2} = 0

forces both squares to vanish. The "circle" has shrunk to zero radius.

Increasing

c

while keeping

g, f

fixed shrinks the circle.

True — since

r = g^{2} + f^{2} - c

, a larger

c

makes

r^{2}

smaller, so the radius shrinks until it hits a point (

r = 0

) and then becomes imaginary. This is exactly the deformation in Figure 2.

3 x^{2} + 3 y^{2} + 6 x + 12 y = 0

is not a circle because it has no constant term.

False —

c = 0

is perfectly allowed; it just means the circle passes through the origin. Dividing by

3

gives

x^{2} + y^{2} + 2 x + 4 y = 0

, a real circle since

g^{2} + f^{2} - c = 1 + 4 - 0 = 5 > 0

The equation

x^{2} + y^{2} + 2 x y + 2 g x + 2 f y + c = 0

can still be a circle for the right

g, f, c

False — any

x y

term rules out a circle. A circle needs equal

x^{2}, y^{2}

coefficients and zero cross term; the

x y

term makes it a rotated conic (see Conic sections general equation, where the full

A x^{2} + B x y + C y^{2} + \dots

family lives and a circle is the case

A = C, B = 0

Spot the error

x^{2} + y^{2} + 8 x - 6 y + 10 = 0

, so centre

= (8, - 6)

and

r^{2} = 10

Two errors: centre is

(- g, - f) = (- 4, 3)

after halving and negating, and

r^{2} = g^{2} + f^{2} - c = 16 + 9 - 10 = 15

, not

10

"For

x^{2} + y^{2} - 4 x + 10 y + 4 = 0

, half of

- 4

- 2

g = - 2

, and the

x

-coordinate of the centre is

- 2

The halving of

g

is right (

g = - 2

), but the centre uses

- g

, so the

x

-coordinate is

- (- 2) = + 2

, not

- 2

(x - 2)^{2} + (y + 5)^{2} = 16

becomes

x^{2} + y^{2} - 4 x + 10 y + 29 = 0

The expansion is right but the constant

16

was never moved across. Setting the right side to zero gives

\dots + 29 - 16 = 0

, i.e.

c = 13

, so the correct form is

x^{2} + y^{2} - 4 x + 10 y + 13 = 0

x^{2} + y^{2} + 2 x + 4 y + 20 = 0

has radius

1 + 4 - 20 = - 15

, so I'll just write

r = 15

Dropping the minus sign hides the real conclusion.

r^{2} = - 15 < 0

means no real circle exists — you cannot rescue it by pretending the number is positive.

"To convert

2 x^{2} + 2 y^{2} + 4 x - 8 y + 2 = 0

I read off

2 g = 4

directly, so

g = 2

You read coefficients before normalising. Divide the whole equation by

2

first (

x^{2} + y^{2} + 2 x - 4 y + 1 = 0

); only then is

2 g = 2 \Rightarrow g = 1

(x - 3)^{2} + (y - 4)^{2} = 25

expands to

x^{2} + y^{2} - 6 x - 8 y + 25 = 0

The constant is wrong. Expanding gives

9 + 16 - 25 = 0

, so

c = 0

: the form is

x^{2} + y^{2} - 6 x - 8 y = 0

. A missing constant is a signal, not a mistake — it means the circle hits the origin.

Why questions

Why does the centre carry a negative of the coefficients,

(- g, - f)

Because the general form writes

+ 2 g x

while the standard form has

- 2 h x

. Matching them term by term gives

2 g = - 2 h

, so

h = - g

(and identically

k = - f

). Figure 1 shows the two expansions lined up so the sign flip is visible, not just asserted.

Why must the

x^{2}

and

y^{2}

coefficients be equal (and ideally

1

Equal coefficients guarantee the same "stretch" in both directions, which is what makes a circle round rather than an ellipse. Concretely, dividing

A x^{2} + A y^{2} + \dots

A

restores unit coefficients; if the two coefficients differed you could not divide them both to

1

, and you'd get an ellipse where the centre/radius formulas fail.

Why is

g^{2} + f^{2} - c

the exact thing we take the square root of?

Because completing the square gives

(x + g)^{2} + (y + f)^{2} = g^{2} + f^{2} - c

— worked in full at the top of this page. The right side is literally

r^{2}

, so its square root is the radius, with no stray factors of

2

or sign errors.

Why can't the constant

c

tell us the radius on its own?

Because

c = h^{2} + k^{2} - r^{2}

blends the centre's distance from the origin (

h^{2} + k^{2}

) with the radius. Example: centre

(3, 4)

with

r = 5

gives

c = 25 + ... - 25 = 0

, while centre

(0, 0)

with

r = 4

gives

c = - 16

— different circles, and

c

alone cannot separate the two effects.

Why does a large positive

c

eventually make the circle imaginary?

Radius squared is

g^{2} + f^{2} - c

; pushing

c

up drains this quantity below zero. When

c = g^{2} + f^{2}

we sit exactly at

r = 0

(a point), and any further increase forces

r^{2} < 0

. Figure 2 shows this as the circle shrinking, pinching to a dot, then disappearing.

Why does completing the square work as a conversion tool here at all?

Because

x^{2} + 2 g x = (x + g)^{2} - g^{2}

is an exact identity — expand the right side and you recover the left. The

(x + g)^{2}

absorbs the linear term, and the leftover

- g^{2}

is the price you pay, revealing the centre offset

- g

. Figure 3 draws this as folding a rectangle of area

2 g x

around a square. It is the algebraic inverse of expanding the Standard form of circle.

Why is checking

g^{2} + f^{2} - c

the very first thing you should do after reading off

g, f, c

Because it decides whether the object even exists as a real circle. There is no point computing a centre for an imaginary circle, and reporting a radius without this check is where most sign-errors survive undetected.

Edge cases

What does

g^{2} + f^{2} - c = 0

represent geometrically?

A point circle: the radius is exactly

0

, so the only real point is the centre

(- g, - f)

. It is the middle frame of Figure 2 — the limiting shape as a circle shrinks to nothing.

What is

x^{2} + y^{2} + 2 x + 4 y + 20 = 0

as a geometric object?

Nothing real —

r^{2} = 1 + 4 - 20 = - 15 < 0

, an imaginary circle with no real points. The equation is true only for complex coordinates (the "past the pinch" frame of Figure 2).

g = f = 0

, what special circle do you get, and where is its centre?

The equation becomes

x^{2} + y^{2} = - c

, a circle centred at the origin

(0, 0)

with

r = - c

. It only exists when

c < 0

; if

c = 0

the origin itself is the single point.

c = 0

, what is special about the circle's position?

It passes through the origin, since

(0, 0)

satisfies

x^{2} + y^{2} + 2 g x + 2 f y + 0 = 0

. The centre and radius are otherwise unaffected: still

(- g, - f)

and

g^{2} + f^{2}

. This is the case that makes Equation of circle from endpoints of diameter problems land on the origin.

Can a real circle have

c

larger than

g^{2} + f^{2}

No — that would force

g^{2} + f^{2} - c < 0

, i.e.

r^{2} < 0

, so no real circle exists. The boundary

c = g^{2} + f^{2}

is exactly the point-circle case.

What happens to centre and radius if you multiply the whole equation by a nonzero constant

λ

Nothing geometrically — the same circle is described. But the coefficients

2 g, 2 f, c

all scale by

λ

, so you must divide back to unit

x^{2}, y^{2}

coefficients before reading off

g, f, c

x^{2} + y^{2} = 0

a circle, a point, or empty?

A point — only

(0, 0)

satisfies it, since here

g = f = c = 0

gives

r^{2} = 0

. It is the degenerate point-circle at the origin.

Once you've confirmed a real circle, what does the general form let you do next?

You can drop straight into Tangent to a circle (using

- g, - f, r

or implicit differentiation) or slide the constant to build a Family of circles

x^{2} + y^{2} + 2 g x + 2 f y + c + λ (\dots) = 0

. Validity must be checked first, or those next steps operate on a non-existent circle.

Connections

Standard form of circle — the clean target $(x - h)^{2} + (y - k)^{2} = r^{2}$ every trap is measured against
Completing the square — the fold that turns linear terms into $(x + g)^{2} - g^{2}$ , the engine behind the negative signs and the $r^{2}$ formula
Equation of circle from endpoints of diameter — a common source of raw general-form equations (often lands with $c = 0$ )
Tangent to a circle — needs a confirmed real circle before tangent lines make sense
Family of circles — sliding the constant/parameter walks a circle toward the point and imaginary boundaries of Figure 2
Conic sections general equation — the full $A x^{2} + B x y + C y^{2} + D x + E y + F = 0$ ; a circle is the special case $A = C, B = 0$